Given one random variable $X$ from a normal distriution with parameteres $\mu=0$ and $\sigma=\theta$. Find critical area of neyman pearson test with $\alpha=0.1$. I know why it is $\{x^2<c\}$ but i know that $c=0.25$. Why is that? $H_0:\theta=4$,$H_1:\theta=1$
$\begingroup$
$\endgroup$
2
-
$\begingroup$ What are the null and the alternative hypotheses? $\endgroup$– V. VancakJun 2, 2017 at 18:09
-
$\begingroup$ my mistake added to the body $\endgroup$– MrowkacalaJun 2, 2017 at 18:14
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
So, according to Neyman-Pearson Lemma, we have to find $k>0$ such that $$ \frac{L(\theta_1;X)}{L(\theta_0;X)} \ge k \, , $$ so reject $H_0$ if $$ \frac{C_{\theta_1}\exp\{-x^2/(2\theta_1)\}}{C_{\theta_0}\exp\{-x^2/(2\theta_0)\}} \propto \exp\{x^2/2(1/\theta_0 - 1/\theta_1)\} = \exp\{ - 15x^2/32\} \ge k. $$ or equivalently if $$ x^2 \le c . $$ Thus, $$ 0.1 = \mathbb{E}_{H_0}I\{x^2 \le c\} = \mathbb{P}(x^2/4 \le c/16) = F_{\chi ^2(1)}(c/16), $$ Finally, $$ c = 16\chi^2(0.1)_{(1)}\approx 0.25. $$
-
$\begingroup$ Is your $C$ independent of $\theta$? We have a Normal distribution here. $\endgroup$– IamKnullJun 18, 2020 at 12:19
-
$\begingroup$ You are right, I've changed the notations to be more precise. But it does not effect the answer. $\endgroup$ Jun 19, 2020 at 2:37